NAS Mechanism - MCAT Organic Chemistry

NAS Mechanism

Updated Apr 17, 2026

Nucleophilic acyl substitution (NAS) is the mechanism by which ALL carboxylic acid derivatives react with nucleophiles. Two phases: addition (nucleophile joins the carbonyl carbon, giving a tetrahedral intermediate) and elimination (the leaving group departs, reforming the carbonyl). Section 8.10 introduced this; here we walk through it in detail with both base and acid catalysis.

Base-Catalyzed (Anionic) NAS

When the nucleophile is already anionic (hydroxide, alkoxide, amide), no separate acid catalyst is needed:

Step 1: Addition. Nu⁻ attacks the carbonyl C. The pi electrons flow onto the oxygen. Product: tetrahedral alkoxide intermediate.

Arrows: Nu⁻ lone pair → C; C=O pi bond → O.

Step 2: Elimination. The oxygen’s lone pair reforms the C=O double bond. The electrons that were in the C-LG bond leave with LG.

Arrows: O’s lone pair → C; C-LG bond → LG.

Net: R-CO-LG + Nu⁻ → R-CO-Nu + LG⁻. One-step-ish mechanism (two arrows, two chemical events, through one stable tetrahedral intermediate).

Acid-Catalyzed (Protonated) NAS

When the nucleophile is neutral (water, alcohol, amine), acid catalysis makes the carbonyl more electrophilic:

Step 1a: Protonate carbonyl oxygen. H⁺ adds to the C=O oxygen.

Step 1b: Neutral nucleophile attacks. The protonated carbonyl is a much stronger electrophile; water, alcohol, or amine can attack its carbon.

Step 2a: Proton shuffle. Proton transfers happen: the new nucleophilic OH/NH gets deprotonated, and then the original leaving group (OH in a carboxylic acid; OR in an ester) gets protonated to become a good leaving group.

Step 2b: Elimination. The protonated LG (now a neutral water or alcohol) leaves with the bond electrons. The C=O reforms.

Step 3: Final deprotonation. The C=O⁺ gets deprotonated to give the neutral product.

This looks complicated but is just Fischer esterification / hydrolysis mechanism from Section 8.8. Each step is a simple proton transfer or attack.

The Tetrahedral Intermediate Is the Branching Point

The tetrahedral intermediate can:

Expel the LG (forward to product): standard NAS.
Expel the Nu back (reverse to starting material): equilibrium situation.
Be protonated and give the stable addition product (in a ketone/aldehyde, which has no LG): this is why aldehydes and ketones give addition not substitution.

Which path wins depends on:

Relative basicity of Nu vs LG: the worse base (= better LG) is expelled preferentially.
Stability of the tetrahedral intermediate.
Solvent and conditions.

Why NAS Requires a Leaving Group

The essential difference between aldehydes/ketones (Ch 6) and carboxylic acid derivatives (Chs 8-9) is the leaving group. Aldehydes and ketones have H or R groups, neither of which can leave. Derivatives have Cl, OR, OCOR’, NR₂, or OH (with acid catalysis) - all of which can leave.

This single structural difference splits the carbonyl world into two:

Aldehydes/ketones: nucleophilic addition only. New Nu stays attached; no group is lost.
Carboxylic acids/derivatives: nucleophilic acyl substitution. Nu replaces LG; net one group in, one group out.

In base-catalyzed hydrolysis of methyl acetate (CH₃COOCH₃), what is the nucleophile, what is the tetrahedral intermediate, and what is the leaving group?

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Nucleophile: OH⁻ (hydroxide). Tetrahedral intermediate: CH₃-C(O⁻)(OCH₃)(OH) - the alkoxide with both -OCH₃ and -OH on the former carbonyl C. Leaving group: CH₃O⁻ (methoxide). The OH⁻ attacks, the tetrahedral intermediate forms, then methoxide leaves and the C=O reforms - giving acetate and methanol. (The final step in saponification is also the fast acid-base deprotonation of the carboxylic acid by hydroxide, which makes the reaction irreversible.)